THERMODYNAMICPROPERTIES ANDCALCULATIONAcademic Resource Center
THERMODYNAMIC PROPERTIESA quantity which is either an attribute of anentire system or is a function of position which iscontinuous and does not vary rapidly overmicroscopic distances, except possibly for abruptchanges at boundaries between phases of thesystem; examples are temperature, pressure,volume, concentration, surface tension, andviscosity. Also known as macroscopic property.
BASIC CONCEPTS-1First Law of Thermodynamic: Although energy assumes many forms, the totalquantity of energy is constant, and when energydisappears in one form it appears simultaneouslyin other forms. (Energy of the system) (Energy ofsurroundings) 0 Ut Q W (nU) Q W dUt dQ dW d(nU) dQ dW There exists a form of energy, known as internalenergy U.
BASIC CONCEPTS-2 PV diagramVirial Equations of StatePV a bP cP2 . Ideal gas: Z 1 or PV RT Van Der Waals Equation of State
For Ideal Gas: Equation for CalculationHeat capacity:dQ dW CvdTdW – PdVdQ CvdT PdVLet V RT/P :
BASIC CONCEPTS-3 Statements of the Second Law: Statement 1: No apparatus can operate in such away that its only effect (in system andsurroundings) is to convert heat absorbed by asystem completely into work done by the system. Statement 2: No process is possible whichconsists solely in the transfer of heat from onetemperature level to a higher one
PRIMARY THERMODYNAMIC PROPERTIES—P, V, T, S & U Combining the first and second laws in reversibleprocess The only requirements are that the system beclosed and that the change occur betweenequilibrium states. H U PV A U – TS G H – TSd(nU) Td(nS) – Pd(nV)d(nH) Td(nS) (nV)dPd(nA) – Pd(nV) – (nS)dd(nG) (nV)dP – (nS)dT
dU TdS – PdVdH TdS VdPdA – PdV – SdTdG VdP – SdTMaxwell’s equation
EXAMPLE 1 Air at 1 bar and 298.15K (25 ) is compressed to 5 barand 298.15K by two different mechanically reversibleprocesses:(a) Cooling at constant pressure followed by heating atconstant volume.(b) Heating at constant volume followed by cooling atconstant pressure.Calculate the heat and work requirements and ΔU andΔH of the air for each path. The following heat capacitiesfor air may be assumed independent of temperature:CV 20.78 and CP 29.10 J mol-1 K-1Assume also for air that PV/T is a constant, regardless ofthe changes it undergoes. At 298.15K and 1 bar the molarvolume of air is 0.02479 m3 mol-1.
SOLUTIONS: In suche case take the system as 1 mol of aircontained in an imaginary piston/cyclinderarrangement. Since the processes considered aremechanically reversible, the piston is imagined tomove in the cylinder withour friction. The finalvolume is(a) During the first step the air is cooled at theconstant pressure of 1 bar until the final volumeof 0.004958 m3 is reached. The temperature ofthe air at the end of this cooling step is:
SOLUTIONS Also,𝑄 𝐻 𝑃𝑉 𝐻 𝑃 𝑉 6,941 1 105 0.004958 0.02479 4,958J During the second step the volume is held constantat while the air is heated to its final state.The complete process represents the sum of itssteps. Hence,Q -6,941 4,958 -1,938JΔU -4,958 4,958 0J Since the first law applies to the entire process, ,and therefore,0 -1,983 WW 1,983J
SOLUTIONS, also applies to the entire process. But and therefore, Hence , and
SOLUTIONS Two different steps are used in this case to reachthe same final state of the air. In the first stepthe air is heated at a constant volume equal to itsinitial valve until the final pressure of 5 bar isreached. The air temperature at the end of thisstep is:For this step the volume is constant, andDuring the second step the air is cooled at theconstant pressure of 5 bar to its final state:
SOLUTIONS Also, 𝑈 𝐻 𝑃𝑉 𝐻 𝑃 𝑉 34,703 5 105 0.004958 0.02479 24,788J For the two steps combined,Q 24,788-34,703 -9,915JΔU 24,788-24,788 0Jand as before During the second step the volume is held constantat V2 while the air is heated to its final state. The property changes and calculated for the givenchange in state are the same for both paths. On theother hand the answers to parts (a) and (b) showthat Q and W depend on the path.
EXAMPLE 2 Air is compressed from an initial condition of 1 bar and25 to a final state of 5 bar and 25 by three differentmechanically reversible processes in a closed system:(a) Heating at constant volume followed by cooling atconstant pressure.(b) Isothermal compression.(c)Adiabatic compression followed by cooling at constantvolume.Assume air to be an ideal gas with the constant heatcapacities, CV (5/2)R and CP (7/2)R. Calculate thework required, heat transferred, and the changes ininternal energy and enthalpy of the air for each process.
SOLUTIONSChoose the system as 1 mol of air, contained inan imaginary frictionless piston/cylinderarrangement. For R 8.314 J mol-1K-1,CV 20.785 CP 29.099J mol-1 K-1 The initial and final conditions of the air areidentical with those of Ex.1, where the molarvolumes are given as:V1 0.02479 V2 0.004958m3 Moreover, since the initial and finaltemperatures are the same, then for all parts ofthe problem:
SOLUTIONS (a) The heat transferred, from Ex.1(b) is Q 9.915J. Thus by the first law applied to the entireprocess:(b) Equation for the isothermal compression ofan ideal gas applies here:(c) The intial adiabatic compression of the airtakes it to its final volume of 0.004958m3. Thetemperature and pressure at this point are:
SOLUTIONSV1T2 T1V2γ 10.02479 0.4 298.15 () 567.57K0.004958V1 γ0.02479 1.4P2 P1 1() 9.52barV20.004958For this step Q 0, andW CV T 20.785 567.57 298.15 5,600JFor the second step at constant V, W 0. For the overall process,W 5,600 0 5,600J
SOLUTIONS Moreover,, and by the first law,Q 𝑈 𝑊 0 5,600 5,600J Although the property changes and are zero foreach process, Q and W are path-dependent. Sincethe work for each of these mechanicallyreversible processes is given by , the work foreach process is proportional to the total areabelow the paths on the PV diagram representingthe process. The relative sizes of these areascorrespond to the numerical values of W.
EXAMPLE 3 Determine the enthalpy and entropy changes ofliquid water for a change of state from 1 bar and25 to 1,000 bar and 50 . The following datafor water are availbale:t/ P/barCP/J mol-1 K- V/cm3 mol- β/K-11125175.30518.071256 10-6251,000.18.012366 10-650175.31418.234458 10-6501,000.18.174568 10-6
SOLUTIONS For application to the change of state described,equations require integration. Since enthalpyand entropy are state functions, the path ofintegration is arbitrary. Since the data indicatethat CP is a weak function of T and that both Vand β are weak functions of P, integration witharithmetic means is satisfactory. The integratedforms of Eqs that result are:
SOLUTIONS For P 1 bar, and for t 50
SOLUTIONS Substitution of these numerical values into theequation for gives:1 513 10 6 323.15 18.204 (1,000 1) H 75.310 323.15 298.15 10cm3 bar J 1 Similarly for ,323.15 513 10 6 18.204 (1,000 1) S 75.310ln 298.1510cm3 bar J 1Note that the effect of a pressure change ofalmost 1,000 bar on the enthalpy and entropy ofliquid water is less than that of a temperaturechange of only 25
THERMODYNAMIC PROPERTIES AND THEIRCHARACTERISTICS
PV diagram Virial Equations . Assume also for air that PV/T is a constant, regardless of the changes it undergoes. At 298.15K and 1 bar the molar volume of air is 0.02479 m3 mol-1. KEYS. SOLUTIONS: